Statistik Bisnis. Week 12 Analysis of Variance

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Statistik Bisnis

Week 12

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Learning Objectives

This week, you learn:

• How to use one-way analysis of variance to

test for differences among the means of

several populations (also referred to as

“groups” in this chapter)

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Chapter Overview

Analysis of Variance (ANOVA) One-Way ANOVA F-test Tukey-Kramer Multiple Comparisons

Levene Test For Homogeneity of Variance Randomized Block Design Tukey Multiple Comparisons Two-Way ANOVA Interaction Effects Tukey Multiple Comparisons

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One-Way Analysis of Variance

• Evaluate the difference among the means of three or more groups

Examples: Accident rates for 1st_{, 2}nd_{, and 3}rd_shift

Expected mileage for five brands of tires

• Assumptions

– Populations are normally distributed – Populations have equal variances

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Hypotheses of One-Way ANOVA

•

– All population means are equal

– i.e., no factor effect (no variation in means among groups)

•

– At least one population mean is different – i.e., there is a factor effect

– Does not mean that all population means are different (some pairs may be the same)

c 3 2 1 0 : μ μ μ μ H      same the are means population the of all Not : H₁

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One-Way ANOVA

The Null Hypothesis is True All Means are the same:

(No Factor Effect)

c 3 2 1 0 :μ μ μ μ H      same the are μ all Not : H₁ _j μ μ μ  

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One-Way ANOVA

The Null Hypothesis is NOT true At least one of the means is different

(Factor Effect is present)

c 3 2 1 0 : μ μ μ μ H      same the are μ all Not : H₁ _j μ μ μ   μ  μ  μ or

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Partitioning the Variation

• Total variation can be split into two parts:

SST = Total Sum of Squares

(Total variation)

SSA = Sum of Squares Among Groups

(Among-group variation)

SSW = Sum of Squares Within Groups

(Within-group variation)

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Partitioning the Variation

Total Variation = the aggregate variation of the individual data values across the various factor levels (SST)

Within-Group Variation = variation that exists among the data values within a particular factor level (SSW)

Among-Group Variation = variation among the factor sample means (SSA)

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Partition of Total Variation

Variation Due to Factor (SSA)

Variation Due to Random Error (SSW)

Total Variation (SST)

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Total Sum of Squares



 





c j n i ij j

X

SST

1 1 2

)

(

Where:

SST = Total sum of squares

c = number of groups or levels

n_j = number of observations in group j X_ij = ith_{observation from group j}

X = grand mean (mean of all data values)

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Total Variation

Group 1 Group 2 Group 3 Response, X

X

2 2 12 2 11

)

(

)

(

)

(

X

SST

c cn

















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Among-Group Variation

Where:

SSA = Sum of squares among groups

c = number of groups

n_j = sample size from group j X_j = sample mean from group j

X = grand mean (mean of all data values)

2 1

)

(

X

n

SSA

j c j j









SST = SSA + SSW

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Among-Group Variation

Variation Due to

Differences Among Groups

i  _j 2 1

)

(

X

n

SSA

j c j j









1 



c

SSA

MSA

Mean Square Among = SSA/degrees of freedom

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Among-Group Variation

Group 1 Group 2 Group 3 Response, X

X

1

X

2 2 2 2 2 2 1 1

(

X

)

n

(

X

)

n

(

X

)

n

SSA















_c c



3

X

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Within-Group Variation

Where:

SSW = Sum of squares within groups

c = number of groups

n_j = sample size from group j X_j = sample mean from group j X_ij = ith_{observation in group j} 2 1 1

)

(

_ij j n i c j

X

SSW

j







 

SST = SSA + SSW

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Within-Group Variation

Summing the variation within each group and then adding

over all groups

n

c

SSW

MSW





Mean Square Within = SSW/degrees of freedom 2 1 1

)

(

_ij j n i c j

X

SSW

j







  j μ

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Within-Group Variation

Group 1 Group 2 Group 3 Response, X 1

X

2 3

X

2 2 2 12 2 1 11

)

(

)

(

)

(

X

_cn

X

c

SSW

c

















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Obtaining the Mean Squares

c

n

SSW

MSW





1 



c

SSA

MSA

1 



n

SST

MST

The Mean Squares are obtained by dividing the various sum of squares by their associated degrees of freedom

Mean Square Among (d.f. = c-1)

Mean Square Within (d.f. = n-c)

Mean Square Total (d.f. = n-1)

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One-Way ANOVA Table

Source of Variation Sum Of Squares Degrees of Freedom Mean Square (Variance) Among Groups c - 1 MSA = Within Groups n - c SSW MSW = Total n – 1 SST SSA MSA MSW F c = number of groups

n = sum of the sample sizes from all groups SSA

c - 1

SSW

n - c

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One-Way ANOVA

F Test Statistic

• Test statistic

MSA is mean squares among groups

MSW is mean squares within groups

• Degrees of freedom

– df₁ = c – 1 (c = number of groups)

– df₂ = n – c (n = sum of sample sizes from all populations)

MSW

MSA

F

_STAT



H₀: μ₁= μ₂ = …_{= μ} c

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Interpreting One-Way ANOVA

F Statistic

• The F statistic is the ratio of the

among

estimate of variance and the

within

estimate of variance

– The ratio must always be positive – df₁ = c -1 will typically be small – df₂ = n - c will typically be large

Decision Rule:

 Reject H₀ if F_STAT > F_α,

otherwise do not reject

H₀ 0



Reject H₀ Do not

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One-Way ANOVA

F Test Example

Anda ingin mengetahui apakah jarak bola yang dipukul oleh tiga stick golf yang berbeda akan

berbeda. Anda memilih lima pemukulan bola secara acak untuk masing-masing stick golf tersebut. Pada tingkat

signifikansi 0,05, apakah terdapat bukti yang

menunjukkan perbedaan pada rata-rata jarak pukulan bola?

Club 1 Club 2 Club 3

254 234 200

263 218 222

241 235 197

237 227 206

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• • • • •

One-Way ANOVA Example: Scatter

Plot

270 260 250 240 230 220 210 200 190 • • • • • • • • • • Distance 1

X

2

X

3

X

227.0 X 205.8 X 226.0 X 249.2 X₁ ₂ ₃    

254 234 200 263 218 222 241 235 197 237 227 206 251 216 204 1 2 3

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One-Way ANOVA Example

Computations

254 234 200 263 218 222 241 235 197 237 227 206 251 216 204 X₁ = 249.2 X₂ = 226.0 X₃ = 205.8 X = 227.0 n₁ = 5 n₂ = 5 n₃ = 5 n = 15 c = 3 SSA = 5 (249.2 – 227)2_{+ 5 (226 – 227)}2_{+ 5 (205.8 – 227)}2_{= 4,716.4} SSW = (254 – 249.2)2_{+ (263 – 249.2)}2₊…_{+ (204 – 205.8)}2_{= 1,119.6} MSA = 4,716.4 / (3-1) = 2,358.2 MSW = 1,119.6 / (15-3) = 93.3 _93.3 25.275 2,358.2 F_STAT  

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One-Way ANOVA Example Solution

H₀: μ₁ = μ₂ = μ₃

H₁: μ_j not all equal

 = 0.05 df₁= 2 df₂ = 12 Test Statistic: Decision: Conclusion: Reject H₀ at  = 0.05

There is evidence that at least one μ_j differs from

the rest 0

 = .05 Reject H₀ Do not reject H 25.275 93.3 2358.2 F_STAT    MSW MSA Critical Value: F_α= 3.89

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The Tukey-Kramer Procedure

• Tells

which

population means are significantly

different

– e.g.: μ₁ = μ₂  μ₃

– Done after rejection of equal means in ANOVA

• Allows paired comparisons

– Compare absolute mean differences with critical range

x

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Tukey-Kramer Critical Range

where:

Q_α = Upper Tail Critical Value from Studentized Range Distribution with c and n - c degrees

of freedom (see appendix E.7 table) MSW = Mean Square Within

n_j and n_j’ = Sample sizes from groups j and j’

















j' j α

n

MSW

Q

ange

Critical R

1

2

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The Tukey-Kramer Procedure:

Example

1. Compute absolute mean differences:

254 234 200 263 218 222 241 235 197 237 227 206 251 216 204 x x 226.0 205.8 20.2 43.4 205.8 249.2 x x 23.2 226.0 249.2 x x 3 2 3 1 2 1            

2. Find the Q_α value from the table in appendix E.7 with

c = 3 and (n – c) = (15 – 3) = 12 degrees of freedom:

3.77

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The Tukey-Kramer Procedure:

Example

5. All of the absolute mean differences are greater than critical range. Therefore there is a significant difference between each pair of means at 5% level of significance.

Thus, with 95% confidence we can conclude that the mean distance for club 1 is greater than club 2 and 3, and club 2 is greater than club 3.

285 16 5 1 5 1 2 3 93 77 3 1 1 2 . . . n n MSW Q ange Critical R j' j α                   

3. Compute Critical Range:

20.2 x x 43.4 x x 23.2 x x 3 2 3 1 2 1       4. Compare:

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ANOVA Assumptions

• Randomness and Independence

– Select random samples from the c groups (or randomly assign the levels)

• Normality

– The sample values for each group are from a normal population

• Homogeneity of Variance

– All populations sampled from have the same variance

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11.7 (cont’d)

The Computer Anxiety Rating Scale (CARS) mengukur level kecemasan komputer individu, dengan skala dari 20 (tidak ada kecemasan) hingga 100 (level tertinggi kecemasan). Peneliti pada Miami University menyebarkan CARS pada 172 mahasiswa bisnis. Salah satu tujuan penelitian ini adalah untuk menentukan apakah terdapat perbedaan tingkat kecemasan komputer pada mahasiswa dengan jurusan yang berbeda. Mereka mendapatkan data sebagai berikut:

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11.7 (cont’d)

Sumber variasi Degrees of Freedom Sum of Squares Mean Squares F Among majors 5 3.172 Within majors 166 21.246 Total 171 24.418 Major n Mean Marketing 19 44,37 Management 11 43,18 Other 14 42,21 Finance 45 41,8 Accountancy 36 37,56

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11.7

a. Lengkapi tabel ringkasan ANOVA diatas.

b. Pada tingkat signifikansi 0,05, apakah terdapat bukti adanya perbedaan pada rata-rata tingkat kecemasan komputer yang dialami oleh mahasiswa dengan jurusan yang berbeda?

c. Jika hasil pada poin (b) menunjukkan ada yang berbeda, gunakan prosedur Tukey-Kramer untuk menentukan jurusan apa yang berbeda tingkat kecemasan komputernya.

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11.11 (cont’d)

Rata-rata jumlah pengunjung pada masing-masing toko dari sebuah peritel ternama (yang memiliki lebih dari 10.000 toko) akhir-akhir ini selalu tetap pada angka 900 orang pengunjung. Untuk meningkatkan jumlah pelanggan, peritel tersebut berencana untuk menurunkan harga kopinya. Peritel tersebut ingin mengetahui seberapa banyak pengurangan harga yang mampu meningkatkan jumlah pengunjung harian tanpa terlalu besar mengurangi keuntungan kasar dari penjualan kopi tersebut.

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11.10 (cont’d)

Sebuah produsen pulpen menyewa jasa agen periklanan untuk membuat iklan produk mereka. Untuk mempersiapkan proyek ini, direktur riset melakukan penelitian mengenai pengaruh iklan pada persepsi produk. Sebuah eksperimen didesain untuk membandingkan lime iklan yang berbeda. Iklan A sangat tidak menonjolkan karakteristik pulpen. Iklan B kurang menonjolkan karakteristik pulpen. Iklan C agak melebih-lebihkan karakteristik pulpen. Iklan D sangat melebih-lebihkan

karakteristik pulpen. Iklan E mencoba

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11.10 (cont’d)

Sebuah sampel yang terdiri dari 30 orang responden diminta untuk mengevaluasi salah satu iklan tersebut (sehingga terdapat 6 responden untuk masing-masing iklan). Setelah membaca iklan dan dapat membayangkan ekspektasi produk dari iklan tersebut, semua responden diberi pulpen yang sama untuk dievaluasi. Responden diizinkan untuk mencoba pulpen tersebut sesuai dengan yang dijanjikan iklan yang mereka terima. Kemudian responden diminta untuk memberi penilaian dari 1 hingga 7 (terendah hingga tertinggi) untuk karakteristik produk tersebut: penampilan, ketahanan, dan kinerja. Total dari ketiga karakteristik dari masing-masing responden dapat dilihat dari tabel berikut: